Question

The $288th$ term of the series $a,b,b,c,c,c,d,d,d,d,e,e,e,e,e,f,f,f,f,f,f,$ is

• A. $u$
• B. $v$
• C. $x$
• D. $w$

Answer 1

Answer: (C)

$x$

The given series is $a,b,b,c,c,c,d,d,d,d,e,e,e,e,e,f,f,f,f,f,f.....$
Here, $a=1,b=2,c=\mathrm{3.............}z=26$
$\therefore$The series will become,
$1,2,3,\mathrm{4.....}$
Now, the no. of terms of series forms the sum of first natural numbers i.e., $\frac{n(n+1)}{2}$
$\therefore \frac{n(n+1)}{2}=288$
$\therefore$ On solving this quadratic equation, it is found that the 23rd term will be the 276th term of series and it represents the letter w.
The first 23 letters will account for the first $\frac{n(n+1)}{2}=276$ terms of series whereas the next letter Y is repeated 24 times.
Sum of first 24 terms$=300$.
This shows that 288th term lies under the sum of sequence 300.
Thus, the 288th term will be 24th letter which is $x$.